The Integral and Kinematics

AP Physics CMrs. Coyle

For Constant Velocity:

Time (s)o

Velocity (m/s)

Area Under Line = Displacement

x = v t

v

t

For a Varying Velocity

v (m/s)

t(s)

For a very small t, each displacement x = vavgtis the small area under curve over that small t.

The total area under the curve from ti to tf is equal to the sum of all the rectangles from ti to tf. It is written as:

vxn = instantaneous velocity, because the t0

vx (t)= velocity as a function of time

IntegralLimit or

When the limits of integration are known the integral is called a Definite Integral

Integral is also known as antiderivative. Velocity is the derivative of displacement with

respect to time. v = dxdt Displacement is the integral (antiderivative)

of velocity as a function of time.

dx=v dt ∆x= ∫dx = ∫v dttixi

xf tf

The definite integral of a polynomial:

∫xn dx = xn+1 | n+1

Where n ≠ -1

xf

xi

xf

xi

Example 1 For the following velocity expressions find the

displacement from 2sec to 5sec. (v is in m/s, t in sec).

a) v(t)= 4t b) v(t)= 3 c) v(t)= t2 + 3 d) v(t)=t-3 +1

Ans: a) 42m, b) 9m, c) 48m, d) 3.105m

Integral is also known as antiderivative.

Acceleration is the derivative of velocity with respect to time.

Change in Velocity is the integral (antiderivative) of acceleration as a function of time.

Example 2:The velocity of an object is given by the relationship v= 3t2 +4.

a) Find the expression for the acceleration in terms of time.

b) Find the expression for the displacement in terms of time.

c) Find the displacement between 2s and 4s.

d) Answer for d) 64m

Example 3: The acceleration of an object is given

by the function a= 6t3 –t2. Find the expression for velocity as a

function of time. Find the change in velocity between 0.2s and 2s.

Answer: 21.3 m/s

Indefinite integral:

n ≠ -1 c is a constant

Example 4 –Evaluating the integral constant The acceleration of a particle along the x axis is

a(t)=7.0t, (t is in sec and a is in m/s2). At t=3s, the velocity is +20m/s. What is the velocity at t=5s?

Answer: 76m/s

Prob. # 56 Separating variables in order to integrate. The acceleration of a marble in a certain fluid is

given by a=-3.00v2 for v>0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble’s speed is reduced to half of its initial value?

Strategy: a=dv/dt

-3.00v2 =dv/dt

Separate the variables on each side:

-3 dt= v-2dt , now integrate and then solve.Ans:0.222s